The Fencing Problem.

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Introduction

Maths Coursework - The Fencing Problem

Maths Coursework

The Fencing Problem

Aim:

A farmer has brought 1000 meters of fencing. With this fencing he wants to enclose an area of land. The farmer wants the fencing to enclose an area of the biggest size. I will investigate different shapes the fencing can make to achieve the largest area.

I am going to start investigating different shape rectangles because they are the easiest shapes to work with, the perimeter all of these shapes will have to be 1000 meters. Below are two rectangles (to scale) showing how different shapes with the same perimeter can have different areas. I will use a scale of 1cm: 100m.

1) 2)

400m (4cm)

Height 300m (3cm)

200m (2cm)

100m (1cm)

Width

I will work out the area of both rectangles by using the formula below. Both rectangles have a perimeter of 1000m

1)

Area of rectangle = height * width

Area of rectangle = 400m * 100m

Area of rectangle = 40000 m²

2)

Area of rectangle = height * width

Area of rectangle = 300m * 200m

Area of rectangle = 60000m²

As you can notice the areas of the rectangles 1 and 2 are different though the perimeters of both are 1000m.

The highlighted row gives the biggest area; after I go past this row I start to repeat my self.

A square with the perimeter 1000m gives me the largest area. I will further investigate this because I was going up in increments of 10. I will go into the decimal widths and lengths.

Below is a table of results.

Width

Length

Area

249

251

62499

249.5

250.5

62499.8

24975

250.25

6249994

250

250

62500

250.25

249.75

62499.9

250.5

249.5

62499.8

251

249

62499

From this table I can see that the perfect square with the perimeter 1000m produces the largest area.

From this graph I can see that the highest point is on the 250m mark. The graph is a parabola and is symmetrical. At the centre point of the graph I reach the highest value for the area, after this I just repeat my self.

In a rectangle, any 2 different length sides will add up to 500, because each side has an opposite with the same length. Therefore in a rectangle of 100m * 400m, there are two sides opposite each other that are 100m long and 2 sides next to them that are opposite each other that are 400m long. This means that you can work out the area if you only have the length of one side.

To find the length of the base of a segment I would divide 1000 by the number of sides, so I could use 1000/s (s=number of sides), but because I want to find half of one segment I use (1000/s)/2. To find the interior angle of the n sided polygon I can use, (360/s)/2. I am finding half of the interior angle of a segment. Then I have to find the length of the perpendicular in the segment. To do this I use the tangent formula. Adjacent = opposite/ Tangent

A = [(1000/n)/2]/ Tan (360/s)/2

Then to find the area of the regular polygon I have to multiply the value of the line A by half of the base (1000/s)/2 then I have to multiply that by the number of sided the regular polygon has. So basically it’s A * base *0.5 * number of sides.

As the table above shows as the number of sides go up the area goes up. Now I am going to investigate the area of a perfect circle. I predict that this will ultimately have the largest area because it has an infinite amount of sides.

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